I know that $\det(M) = \prod_{k} \lambda_{k}$ if $\{\lambda_{k}\}_{k=1}^{n}$ are the eigenvalues of $M$ (complex matrix), repeated according to algebraic multiplicity. In this case, if $2, 3, 4, 5$ are the only eigenvalues of $M$, it would seem that $\det(M) > 120$ when $n > 4$ ?
The question I am looking at assumes $2, 3, 4, 5$ to be the eigenvalues of $M$ and $120$ to be the value of the determinant. But it doesn't say $M$ is $4\times 4$, only $n\times n$. It's not that this is wrong, but it seems like a typo. It asks if $M$ is diagonalizable (which if I am right that $M$ must be $4\times 4$ then it is). Thanks.
As you observed if $n >4$ then $\det(M) \neq 120$. This gives that $n=4$, and hence $M$ is diagonalizable.
To make the first observation more precise, denote by $n_\lambda$ the algebraic multiplicity of $\lambda$. Then $n_\lambda \geq 1$. You then have
$$120=\det(M)= 2^{\lambda_2}\cdot 3^{\lambda_3}\cdot 4^{\lambda_4}\cdot 5^{\lambda_5} = 120\cdot 2^{\lambda_2-1}\cdot 3^{\lambda_3-1}\cdot 4^{\lambda_4-1}\cdot 5^{\lambda_5-1}$$ and $$2^{\lambda_2-1}\cdot 3^{\lambda_3-1}\cdot 4^{\lambda_4-1}\cdot 5^{\lambda_5-1} \geq 1$$ since $\lambda_j-1 \geq 0$. You can only get equality if $\lambda_j=1$ for all $j=2,3,4,5$.